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Moments of inertia

  1. Apr 11, 2005 #1
    Hi, can anyone help me understand how to find the moments of inertia for the following:
    1- A triangular lamina (isosceles) of mass M, base 2B and height H. about line of symmetry.
    2- A uniform lamina of mass M, bounded by the curve with equation y²=4ax and the line x = 4a about the x-axis.

    For (1), I managed to get the answer, but I’m not sure if my way is right: it was finding M.I. of a rectangle base 2B, height H, mass 2M about the line of symmetry through the base, and divide it by 2 (as the triangle is the half of the rectangle!) I just don’t know if this method is right or wrong, or whether there is another method that I should had used. Those questions are from the book, and the book doesn’t explain how to find M.I. for such shapes, it only shows: rod, hoop, and discs. But not a triangle or curves!

    Thank you very much,
    Help is appreciated
     
  2. jcsd
  3. Apr 11, 2005 #2

    arildno

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    You are proceeding along a wrong track here, I'm afraid.
    Have you learnt that in general, an object's moment of inertia I with respect to an axis is given by the integral:
    [tex]I=\int_{V}r^{2}dm[/tex]
    where V is the volume of the object (in your 2-D case, an area), r the distance of a mass point within the object to the axis, and dm the (infinitesemal) mass of the mass point?
     
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